Disc deceleration with flexible shaft 3

You will get a complex equation linking the torque to time. At any instant the change in rate will be due to the torque, but picture locking the far end of the shaft - the disc will slow, stop, and with the energy stored in the shaft, start to rotate the opposite direction, oscillating with some frequency and with the torque oscillating with it.

If you ignore the inertia of the shaft, torque is the same at both ends of the shaft and disc acceleration is always going to be Torque/Inertia.

(It only becomes complex if you need to compute the motion of the disc in response to an angular displacement function at the drive end.)

je suis charlie

Thanks.
My instinct is that the decel will be softened by the flexible shaft, similar to how a thick sole on a trainer reduces the force by increasing the impact time. I don't know how to calculate it though.
Alternatively, perhaps it just introduces a lag.

Even with a thick sole the force on the bottom of the foot is the same as the force on the pavement.

What changes is the time vs. displacement and force, just as in this case. The time is extended so the force peak is decreased.

Instead of the rate of acceleration (deceleration = acceleration with a negative sign) making a step change at the moment that the braking force is applied, it will ramp up as the shaft twists, with the flexible end first overshooting the deceleration and then catching up. If the braking force is sufficient to momentarily stop the flexible end of the shaft, you'll get some really interesting and potentially noisy and violent stick-slip phenomena as the flexible end of the shaft first stops then the shaft twists enough for the torque to overcome the static coefficient of friction which then violently lets go. Presumably this connecting shaft is going to have a non-zero mass, and will thus have a resonant frequency, first overshooting the initial braking position and then stopping again, then twisting-up and breaking free again.

what are you building ?

The problem as stated is a very simple one.
[ol 1]
[li]The torque at the braking end is known.[/li]
[li]There is no mention of friction or stick-slip effects.[/li]
[li]The disc has a high MOI so the shaft MOI can be neglected.[/li]
[/ol]
I repeat - if the torque is known, the motion of the disc is easy to compute.

je suis charlie

Using D for d/dt and a for angle J*DDa+c*Da+k*a+T(t)=0

where T is the torque on the end of the shaft. You can correct J and k for the additional effect of the shaft but that is usually neglected. c is also often neglected - there isn't much damping in typical shaft flywheel systems. That of course will result in an undamped SDOF system which will oscillate forever.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

Thanks Greg,

I take it DDa is the acceleration of the disc/shaft system.

Thus, ignoring damping and assuming T is constant, DDa = (-T - k.a)/J.

Using this in an angle stepping spreadsheet, the accel is linearly proportional to the angle and just keeps increasing (negatively), but I can limit it to the accel given if the shaft were rigid i.e. DDa_min = -T/J (unless the speed reaches zero first which I can calculate using v[sup]2[/sup] = v[sub]0[/sub][sup]2[/sup] + 2.DDa.a).

Am I on the right track?

Yes if you apply a constant torque the disc will acclerate at a linear rtae, without limit. Confusion may arise when braking, since a braking force using friction is a very weird function in ODE terms, it is a unit step opposing the current motion of the end of the shaft.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

Flexible how?

Flexible in torsion?

MintJulep said:

Flexible in torsion?

Click to expand...

Yes